Codeforces 718C solution

　　　　　　　　　　　　　　　　　　　　　　　　　C. Sasha and Array

 

　　　　　　　　　　　　　　　　　　　　　　　time limit per test : 

5 seconds

　　　　　　　　　　　　　　　　　　　　　memory limit per test : 

256 megabytes

Description

Sasha has an array of integers a₁, a₂, ..., a_n. You have to perform m queries. There might be queries of two types:

1. 1 l r x — increase all integers on the segment from l to r by values x;
2. 2 l r — find , where f(x) is the x-th Fibonacci number. As this number may be large, you only have to find it modulo 10⁹ + 7.

In this problem we define Fibonacci numbers as follows: f(1) = 1, f(2) = 1, f(x) = f(x - 1) + f(x - 2) for all x > 2.

Sasha is a very talented boy and he managed to perform all queries in five seconds. Will you be able to write the program that performs as well as Sasha?

Input

The first line of the input contains two integers n and m (1 ≤ n ≤ 100 000, 1 ≤ m ≤ 100 000) — the number of elements in the array and the number of queries respectively.

The next line contains n integers a₁, a₂, ..., a_n (1 ≤ a_i ≤ 10⁹).

Then follow m lines with queries descriptions. Each of them contains integers tp_i, l_i, r_i and may be x_i (1 ≤ tp_i ≤ 2, 1 ≤ l_i ≤ r_i ≤ n, 1 ≤ x_i ≤ 10⁹). Here tp_i = 1 corresponds to the queries of the first type and tp_i corresponds to the queries of the second type.

It's guaranteed that the input will contains at least one query of the second type.

Output

For each query of the second type print the answer modulo 10⁹ + 7.

Example Input

5 4
1 1 2 1 1
2 1 5
1 2 4 2
2 2 4
2 1 5

Example Output

5
7
9

Note

Initially, array a is equal to 1, 1, 2, 1, 1.

The answer for the first query of the second type is f(1) + f(1) + f(2) + f(1) + f(1) = 1 + 1 + 1 + 1 + 1 = 5.

After the query 1 2 4 2 array a is equal to 1, 3, 4, 3, 1.

The answer for the second query of the second type is f(3) + f(4) + f(3) = 2 + 3 + 2 = 7.

The answer for the third query of the second type is f(1) + f(3) + f(4) + f(3) + f(1) = 1 + 2 + 3 + 2 + 1 = 9.

---

没什么可说的，就是线段树维护fib递推矩阵，线段树不难写，难写的是矩阵部分，细节很多。要是封装性太强容易把代码写得很java。

 /* basic header */
 #include <iostream>
 #include <cstdio>
 #include <cstdlib>
 #include <string>
 #include <cstring>
 #include <cmath>
 #include <cstdint>
 #include <climits>
 #include <float.h>
 /* STL */
 #include <vector>
 #include <set>
 #include <map>
 #include <queue>
 #include <stack>
 #include <algorithm>
 #include <array>
 #include <iterator>
 /* define */
 #define ll long long
 #define dou double
 #define pb emplace_back
 #define mp make_pair
 #define fir first
 #define sec second
 #define sot(a,b) sort(a+1,a+1+b)
 #define rep1(i,a,b) for(int i=a;i<=b;++i)
 #define rep0(i,a,b) for(int i=a;i<b;++i)
 #define repa(i,a) for(auto &i:a)
 #define eps 1e-8
 #define int_inf 0x3f3f3f3f
 #define ll_inf 0x7f7f7f7f7f7f7f7f
 #define lson curPos<<1
 #define rson curPos<<1|1
 /* namespace */
 using namespace std;
 /* header end */

 const int maxn = 1e5 + ;
 const int mod = 1e9 + ;

 struct Matrix
 {
     ll data[][];
     Matrix()
     {
         data[][] = data[][] = data[][] = data[][] = ;
     }
     //变为转移矩阵
     inline void tran()
     {
         data[][] = data[][] = data[][] = , data[][] = ;
     }
     //初始化为对角矩阵
     inline void init()
     {
         *this = Matrix();
         rep1(i, , ) data[i][i] = ;
     }
     //返回矩阵第x行
     inline ll *operator[](int x)
     {
         return data[x];
     }
     //定义矩阵乘法
     inline void operator*=(Matrix &rhs)
     {
         Matrix qAns;
         rep1(k, , )
         {
             rep1(i, , )
             {
                 rep1(j, , )
                 qAns[j][i] = (qAns[j][i] + data[j][k] * rhs[k][i]) % mod;
             }
         }
         *this = qAns;
     }
     //矩阵快速幂
     inline void operator^=(int x)
     {
         Matrix base = *this, qAns;
         rep1(i, , ) qAns[i][i] = ;
         for (register int i = x; i; i >>= , base *= base)
             if (i & ) qAns *= base;
         *this = qAns;
     }
     inline void operator+=(Matrix &rhs)
     {
         rep1(i, , )
         {
             rep1(j, , )
             {
                 data[i][j] = (data[i][j] + rhs[i][j]) % mod;
             }
         }
     }
     //输出
     inline void pr()
     {
         rep1(t, , )
         {
             rep1(i, , ) printf("%d ", data[t][i]);
             puts("");
         }
     }
 } seg[maxn << ], add[maxn << ];

 int n, m;
 Matrix qAns, markdown;
 bool lazyTag[maxn << ];

 inline void pushdown(int curPos)
 {
     if (!lazyTag[curPos]) return;
     seg[lson] *= add[curPos]; seg[rson] *= add[curPos];
     add[lson] *= add[curPos]; add[rson] *= add[curPos];
     lazyTag[lson] = lazyTag[rson] = ;
     add[curPos].init(); lazyTag[curPos] = ;
 }

 void build(int curPos, int curL, int curR)
 {
     if (curL == curR)
     {
         seg[curPos].tran();
         add[curPos].init();
         int tmp; scanf("%d", &tmp);
         seg[curPos] ^= tmp - ;
         return;
     }
     int mid = (curL + curR) >> ;
     add[curPos].init();
     build(lson, curL, mid); build(rson, mid + , curR);
     seg[curPos] = Matrix();
     seg[curPos] += seg[lson]; seg[curPos] += seg[rson];
 }

 void update(int curPos, int curL, int curR, int qL, int qR)
 {
     if (qL <= curL && curR <= qR)
     {
         add[curPos] *= markdown; seg[curPos] *= markdown; lazyTag[curPos] = ;
         return;
     }
     int mid = (curL + curR) >> ;
     pushdown(curPos);
     if (qL <= mid) update(lson, curL, mid, qL, qR);
     if (mid < qR) update(rson, mid + , curR, qL, qR);
     seg[curPos] = Matrix();
     seg[curPos] += seg[lson]; seg[curPos] += seg[rson];
 }

 void query(int curPos, int curL, int curR, int qL, int qR)
 {
     if (qL <= curL && curR <= qR)
     {
         qAns += seg[curPos];
         return;
     }
     int mid = (curL + curR) >> ;
     pushdown(curPos);
     if (qL <= mid) query(lson, curL, mid, qL, qR);
     if (mid < qR) query(rson, mid + , curR, qL, qR);
     seg[curPos] = Matrix();
     seg[curPos] += seg[lson]; seg[curPos] += seg[rson];
 }

 int main()
 {
     scanf("%d%d", &n, &m);
     build(, , n);
     rep1(cnt, , m)
     {
         int op; scanf("%d", &op);
         if (op == )
         {
             int x, y, t; scanf("%d%d%d", &x, &y, &t);
             markdown.tran(); markdown ^= t;
             update(, , n, x, y);
         }
         else
         {
             int x, y; scanf("%d%d", &x, &y);
             qAns = Matrix();
             query(, , n, x, y);
             printf("%lld\n", (qAns[][] + qAns[][]) % mod);
         }
     }
     return ;
 }